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Chapter 9: Problem 1216

At \(100 \mathrm{~K}\) and \(0.1\) atmospheric pressure, the volume helium gas is 10 liters. If volume and pressure are doubled, its temperature will change to (A) \(127 \mathrm{~K}\) (B) \(400 \mathrm{~K}\) (C) \(25 \mathrm{~K}\) (D) \(200 \mathrm{~K}\)

Short Answer

Expert verified
The temperature of helium gas will change to \(200 \,\text{K}\) when the volume and pressure are doubled. The correct answer is (D) \(200 \,\text{K}\).

Step by step solution

01

Write down the Ideal Gas Law formula

We will use the Ideal Gas Law formula, which is represented by: \(PV = nRT\) Here, P = pressure, V = volume, n = amount of gas in moles, R = gas constant, and T = temperature in kelvin.
02

Analyze the given information

We are given the initial volume (\(V_1\)), pressure (\(P_1\)), and temperature (\(T_1\)) of the helium gas. We are also told that both the volume and pressure are doubled, which results in a change in the temperature (\(T_2\)). Therefore, we need to find the value of \(T_2\). Initial conditions: \(P_1=0.1\) atm, \(V_1=10\) L, \(T_1=100\) K. Final conditions: \(P_2=2P_1\), \(V_2=2V_1\), \(T_2=\) unknown.
03

Convert the given pressure to the same units

The units of pressure should be consistent within the formula. The Gas Constant (R) used in the given exercise is \(8.314 \,\text{J/molK}\), thus we need to convert the given pressure from atm to Pa (Pascals). The conversion factor is as follows: 1 atm = 101325 Pa. Now we can convert the given pressure: \(P_1 = 0.1 \,\text{atm} \times \frac{101325 \,\text{Pa}}{1 \,\text{atm}} = 10132.5 \,\text{Pa}\)
04

Apply the Ideal Gas Law formula for the initial condition

Now, let's use the Ideal Gas Law to find the unknown quantity of moles (n) in the helium gas. Rearranging the formula for calculating n, we get: \(n = \frac{P_1V_1}{RT_1}\). Substituting the given values: \(n = \frac{10132.5 \,\text{Pa} \times 10 \,\text{L}}{8.314 \,\text{J/molK} \times 100 \,\text{K}}\) Note: 1 L = 0.001 m³, we need to convert L to m³: \(n= \frac{10132.5 \,\text{Pa} \times 0.01 \,\text{m}^3}{8.314 \,\text{J/molK} \times 100 \,\text{K}}\)
05

Solve for n

Calculate the value of n: \(n \approx 0.122 \,\text{moles}\)
06

Calculate the final conditions and apply the Ideal Gas Law formula

We know that the pressure and volume are doubled, so we have the final conditions: \(P_2 = 2P_1 = 20265 \,\text{Pa}\) \(V_2 = 2V_1 = 0.02 \,\text{m}^3\) Now we apply the Ideal Gas Law formula again for the final conditions: \(P_2V_2 = nRT_2\) Rearranging the formula for calculating \(T_2\), we get: \(T_2 = \frac{P_2V_2}{nR}\). Substituting the values: \(T_2 = \frac{20265 \,\text{Pa} \times 0.02 \,\text{m}^3}{0.122 \,\text{moles} \times 8.314 \,\text{J/molK}}\)
07

Solve for T_2

Calculate the value of T_2: \(T_2 \approx 200 \,\text{K}\) The temperature of helium gas will change to 200 K when the volume and pressure are doubled. Therefore, the correct answer is (D) \(200 \,\text{K}\).

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Most popular questions from this chapter

The temperature of an ideal gas is increased from \(27^{\circ} \mathrm{C}\) to \(127^{\circ} \mathrm{C}\), then percentage increase in \(\mathrm{v}_{\mathrm{rms}}\) is (A) \(33 \%\) (B) \(11 \%\) (C) \(15.5 \%\) (D) \(37 \%\)

The rms speed of the molecules of a gas at a pressure \(10^{5} \mathrm{~Pa}\) and temperature \(0^{\circ} \mathrm{C}\) is \(0.5 \mathrm{~km} / \mathrm{s}\). If the pressure is kept constant but temperature is raised to $819^{\circ} \mathrm{C}$, the rms speed becomes (A) \(1.5 \mathrm{kms}^{-1}\) (B) \(2 \mathrm{kms}^{-1}\) (C) \(1 \mathrm{kms}^{-1}\) (D) \(5 \mathrm{kms}^{-1}\)

When the pressure on \(1200 \mathrm{ml}\) of a gas is increased from $70 \mathrm{~cm}\( to \)120 \mathrm{~cm}$ of mercury at constant temperature, the new volume of the gas will be (A) \(400 \mathrm{ml}\) (B) \(600 \mathrm{ml}\) (C) \(700 \mathrm{ml}\) (D) \(500 \mathrm{ml}\)

The rms speed of a gas at a certain temperature is \(\sqrt{2}\) times than that of the Oxygen molecule at that temperature, the gas is (A) \(\mathrm{SO}_{2}\) (B) \(\mathrm{CH}_{4}\) (C) \(\mathrm{H}_{2}\) (D) \(\mathrm{He}\)

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